ME 4133 Chapter : Kin Sol Hw4
4.1)
2
O
A
B
C
O4
30°
90°
2
3
4
srad /2
2=
ω
srad /4
2=
ω
&
2
O
A
B
C
O4
30°
90°
2
3
4
srad /2
2=
ω
srad /4
2=
ω
&
Given:
O2A
=
15in
inBO 10
4
=
inAC 10
=
inOO 30
42
=
inAB 2.17
=
(From figure)
Find v and
c
ω
3 graphically.
The velocity of point
B
can be found in terms of velocity of point using the equation
v
B
=vA+
ω
3×r
B
/A (1)
The magnitude of the velocity of point
A
is sinAOvA/30
22 ==
ω
and its direction is
perpendicular to link 2. Similarly, the velocity of point
B
is perpendicular to link 4 and the
direction of
ω
3×r
B
/A is perpendicular to
A
B.
Procedure to draw the velocity diagram (Figure 1(a)):
- both in magnitude and direction are known
vA
- only direction is known and is perpendicular to link 4 vB
ω
3×rB/
A
- only direction is known and is perpendicular to
A
B
Figure 1(a) Figure 1(b)
Scale used for velocity diagram 1 unit=1 in/sec
1815.11
v
O
A
30
A
v
AB
r/3 ×
ω
B
v
v
O
259.26
A
v
AC /3 r
×
ω
C
v
C
664.33 BC /3 r
ω
×
From Figure 1(a) and equation (1) and the scale used
sin
AB /259.26
/3 =×r
ω
srad
rAB
AB /5267.1
/
/3
3=
×
=r
ω
ω
To find the velocity of point we use the equation
C
ACAC /3 rvv ×+=
ω
(2)
- both in magnitude and direction are known
vA
AC /3 r×
ω
- both magnitude and direction (perpendicular to ) are known
AC
Then the velocity diagram that represents equation (2) can be drawn and is shown in figure 1(b).
The acceleration of point
B
is given by
()
ABABABABAB //// 2rωvωrωωaaa
×
+
×
+
××++= &
Since link 3 is rigid and 0
/=
AB
v0
/
=
AB
a
()
ABABAB /3/33 rωrωωaa
×
+××+= &.
Points
A
and
B
have circular motion and their accelerations can be split into radial and
tangential components
()
ABAB
T
A
R
A
T
B
R
B/3/33 rωrωωaaaa ×+××++=+ &
where
2
2
2
2/60 sinAOaR
A==
ω
,
2
22 /60 sinAOaT
A==
ω
&,
and
2
4
2
4
2
4/5.12 sin
BO
v
BOa B
R
B===
ω
.
The direction of is perpendicular to link 4,
T
B
a
is from
(
AB /33 rωω ××
)
B
to
A
and is perpendicular to
AB /3 rω×
&
A
B.
Using this the velocity diagram shown in Figure 1(c) is drawn.
R
A
a
T
A
a
A
a
R
B
a
T
B
a
()
AB
r/33 ××
ω
ω
AB
r/3 ×
ω
&
Scale: 1
2
/10 sinunit =
Figure 1(c)
Form the figure
2
/3 /162.29 sinr AB =×
ω
&
2
/3
3/695.1 srad
AB
rAB =
×
=⇒
ω
ω
&
& and the direction is CW.
To find the acceleration of point C, we place the observer at and look at point C. The
acceleration of point is given by
A
C
()
ACACAC /3/33 rωrωωaa
×
+××+= &
and is from to
()
2
/33 /3.23 sin
AC =×× rωω C
A
.
and is perpendicular to .
2
/3 /72.28 sin
AC =×rω
&AC
The acceleration diagram that represents this equation is shown in Figure 1(d).
C
v
A
v
()
AC
r/33 ××
ω
ω
AC
r/3
ω
×
&
Figure (d)
Document Summary
The velocity of point b can be found in terms of velocity of point using the equation v b = v a + 3 rb / a (1) The magnitude of the velocity of point a is and its direction is perpendicular to link 2. Similarly, the velocity of point b is perpendicular to link 4 and the direction of 3 rb / a is perpendicular to ab. Procedure to draw the velocity diagram (figure 1(a)): v a vb. 3 rb / a - only direction is known and is perpendicular to ab. Only direction is known and is perpendicular to link 4. Scale used for velocity diagram 1 unit=1 in/sec vo. From figure 1(a) and equation (1) and the scale used. To find the velocity of point v v r. C we use the equation (2) v a. Both in magnitude and direction are known.
